PE 3 - Time, Speed & Distance | Arithmetic - Time, Speed & Distance

Answer: Option C

Video Explanation

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Explanation :

Burning rate of 10 cm long candle = 4/4 = 1 cm/hr.

Burning rate of 12 cm long candle = 10/5 = 2 cms/hr.

Let the candles be of equal length after t hours.

∴ 10 – t × 1 = 12 – t × 2

⇒ t = 2 hours.

∴ It will takes 2 hours for the two candles to become equal in length.

Hence, option (c).

Answer: Option A

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Explanation :

The speed of two persons is 36 km/h and 25 km/h. 

When 2 people start from opposite ends and to and fro between the two ends, time taken to meet for nth time 

= $\frac{\left(2n-1\right)D}{a+b}$ [Here, ½< a/b < 2]

∴ $(2n-1)\times \frac{60}{36+25}$ ≤ 10

⇒ 2n – 1 ≤ 61/6

⇒ n ≤ 5.6

∴ Both of them will meet 5 times in 10 hours.

Hence, option (a).

Answer: Option D

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Explanation :

The last bus to arrive in Chennai before we leave would’ve arrived at 11:30. This bus would’ve left Bangalore at 8 a.m. We will not be able to meet this bus but will meet other buses which start after this one.

∴ We will meet buses which left Bangalore at 9 a.m. and after that.

If we start at 12 noon, we will reach Bangalore at 3:30 p.m.

∴ We will mee all buses which left till 3 p.m.

⇒ We meet all buses starting from 9 a.m. till 3 p.m. i.e., 7 buses.

Hence, option (d).

Answer: Option D

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Explanation :

Let the distance between the two ant holes be 600d feet.

∴ Speed of first ant = $\frac{600d\times \frac{8}{100}}{3}$ = 16d feet/hr

∴ Speed of second ant = $\frac{600d\times \frac{7}{120}}{2.5}$ = 14d feet/hr

Till meeting point first ant travels 800 ft and second ant travels 600d – 800 ft.

∴ $\frac{800}{600d-800}=\frac{16d}{14d}$

⇒ 700 = 600d – 800

⇒ d = 2.5

∴ Speed of second ant and = 14 × 2.5 = 35 feet/hr

Hence, option (d).

Answer: Option C

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Explanation :

Accident occurs after travelling 100 kms. Let the remaining distance to be travelled is d kms.

Train gets late by 4 hours in travelling d kms at reduced speed.

Also, train will get late by 2 hours in travelling d – 50 kms at reduced speed.

⇒ Train will get late by 2 hours in travelling 50 kms at reduced speed.

Since, train gets late by same time while travelling 50 or d – 50 kms.

⇒ 50 = d – 50

⇒ d = 100 kms.

∴ The total distance to be travelled by train = 100 + 100 = 200 kms.

Alternately,

Accident occurs after travelling 100 kms. Let the remaining distance to be travelled is d kms.

Since speed become 2/3rd it would take 3/2 times the normal time.

Let the normal time taken to travel d kms is t. Since the train gets delayed by 2 hours,

∴ 3t/2 – t = 2

⇒ t = 4 hours.

∴ It takes 4 hours to travel d kms   …(1)

Now, had the accident occurred after 50 kms, the distance yet to travel would have been d – 50 kms.

Train gets late by 1 hour during this journey.

∴ 3t’/2 – t’ = 1

⇒ t’ = 2 hours.

∴ It takes 2 hours to travel d – 50 kms   …(2)

From (1) and (2) we see that train takes 2 hours to travel 50 kms.

∴ Normal speed of the train = 50/2 = 25 kmph

Now, train travels d kms in 4 hours

∴ d = 25 × 4 = 100 kms.

The total distance to be travelled by train = 100 + 100 = 200 kms.

Hence, option (c).

Answer: Option B

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Explanation :

Let speed of A = a and that of B = b.

When B  has run 200 meters A and B will be 800 meters apart.

Now when they meet A has run 500 meters hence B would have run 800 – 500 = 300 meters.

∴ Ratio of speeds of A and B = 500 : 300 = 5 : 3

They meet at a distance of 500 meters from the starting point i.e., at a point exactly opposite the starting point.

Now for them to come at the starting point both of them have to travel 500 meters. Since A’s speed is greater than b’s speed A will cover 500 meters before B and in the same time B will cover only 300 meters.

∴ When A reaches starting point B will be still 200 meters away from the starting point.

Hence, option (b).

Answer: Option D

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Explanation :

Let speeds of A and B be a and b respectively.

In the first heat A gives B a start of 100 meters and beats him by 40 seconds

∴ $\frac{1000}{a}$ + 40 = 900/b   …(1)

In the second heat A gives B a start of 20 seconds and is beaten by 300 meters

∴ $\frac{1000}{a}$ + 20 = 700/b   …(2)

Solving (1) and (2), we get, 

a = 20 m/s and b = 10 m/s

∴ Time taken by them to complete 1 km is

A = 1000/20 = 50 seconds
B = 1000/10 = 100 seconds

Hence, option (d).

Answer: Option B

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Explanation :

Let the speed of the faster person be x kmph, then speed of the slower person be (x - 2) km/h.

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Distance travelled by them in 2 hours will be, 2x and 2(x - 2) kms.

∴ (2x)² + (2(x - 2))² = 20²

⇒ 8x² – 16x + 16 = 400

⇒ 8x² – 16x - 384 = 0

⇒ x² – 2x - 48 = 0

⇒ x = 8 or -6. (-6 will be rejected)

∴ Speed of the faster person is 8 kmph.

Hence, option (b).

Answer: Option D

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Explanation :

Let speed of boatman in still water be b and speed of current = r.

Initially, 

Speed of boat against the current = b – r

Speed of boat with the current = b + r

Since the boat covers twice the distance downstream as compared to upstream in the same time, it means downstream speed is twice that of upstream speed.

∴ b + r = 2(b - r)

⇒ b = 3r   …(1)

Now, when speed of current is increased by 2 kmph, upstream speed is 5 times downstream speed.

∴ (b + r + 2) = 5(b – (r + 2))

⇒ 4r + 2 = 5(2r - 2)

⇒ r = 2 and b = 6 kmph.

Hence, option (d).

Answer: Option A

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Explanation :

Man could see the car for 3 minutes and up to a distance of 300 m.

∴ Relative speed of car w.r.t. man = 0.3/(1/60) = 18 kmph.

⇒ Relative speed = 18 = S_car – S_man = S_car – 2

⇒ S_car = 20 kmph.

Hence, option (a).